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User blog:B1mb0w/Strong D Function
Strong D Function The strong D function is based on a weaker deeply nested Ackermann function called d. The rules are similar with the significant change being that the D function: \(D(x_1,x_2,x_3,x_4,...,x_n)\) expands to this function: \(D( x_1-1, D(x_1,x_2,x_3,x_4,...,x_n-1), ..., D(x_1,x_2,x_3,x_4,...,x_n-1))\) The same expansion is used to replace each input parameter \(x_2\) to \(x_n\). Also refer to the Leading and Trailing Zero rules: L1 and T1, for more information. For 2 parameters, the D function is equivalent to the d function: \(d(a,b)=d(a-1,d(a,b-1))=D(a,b)=D(a-1,D(a,b-1))\) For 3 parameters, the D function quickly dominates the weaker d function: \(d(a,b,c)=d(a-1,d(a,b-1),d(a,b,c-1))\) \(D(a,b,c)=D(a-1,D(a,b,c-1),D(a,b,c-1))\) Calculated Examples \(D() = 0\) This is a null function that always returns zero. \(D(3) = 4\) This is the successor function \(D(1,2) = 5\) This is the same as d(1,2) \(D(1,0,0)\) expands as follows: \(= D(0, D(0,1,1), D(0,1,1)) = D(4,4) = d(4,4)\) Using the Comparison Rule C1 \(d(m,n) >> f_{m-1}(n+2)\) we get \(>> f_3(6) >> f_{\omega}(3)\) \(D(1,0,1)\) expands as follows: \(= D(0, D(1,0,0), D(1,0,0)) = D(D(4,4),D(4,4)) >> D(f_{\omega}(3),f_{\omega}(3))\) \(>> f_{\omega}(f_{\omega}(3)) = f_{\omega}^2(3)\) My calculations show that \(D(1,0,n) >> f_{\omega}^{n+1}(3)\) e.g. \(D(1,0,2) >> f_{\omega}^3(3) >> f_{\omega+1}(3)\) and \(D(1,0,n) >> f_{\omega}^{n-2}(f_{\omega+1}(3))\) when n>2 More Examples with 3 parameters \(D(1,1,0) = D(0,D(1,0,1),D(1,0,1))\) which is equal to \(D(1,0,2) >> f_{\omega+1}(3)\) Similarly \(D(1,1,1) = D(0,D(1,0,2),D(1,0,2)) = D(1,0,3) >> f_{\omega}^{1}(f_{\omega+1}(3))\) and \(D(1,1,n) >> f_{\omega}^{n}(f_{\omega+1}(3))\) Next \(D(1,2,0) = D(0,D(1,1,2),D(1,1,2)) = D(1,1,3) >> f_{\omega}^{3}(f_{\omega+1}(3))\) \(D(1,2,1) = D(0,D(1,1,3),D(1,1,3)) = D(1,1,4) >> f_{\omega}^{4}(f_{\omega+1}(3))\) \(D(1,2,n) >> f_{\omega}^{n+3}(f_{\omega+1}(3))\) and \(D(1,3,n) >> f_{\omega}^{n+7}(f_{\omega+1}(3))\) and \(D(1,m,n) >> f_{\omega}^{n-3+(m+2).(m+1)/2)}(f_{\omega+1}(3))\) e.g. \(D(1,2,2) >> f_{\omega}^{5}(f_{\omega+1}(3))\) D function examples with 3 parameters - continues \(D(2,0,0)\) grows significantly faster \(= D(1,D(1,2,2),D(1,2,2))\) and \(>> f_{\omega}^{m-3+(m+2).(m+1)/2)}(f_{\omega+1}(3))\) where \(m >> f_{\omega}^{5}(f_{\omega+1}(3))\) Using a general rule G1 \(f_{\omega}^{f_b^c(a).2}(a) >> f_b^{c+1}(a)\) we get \(f_{\omega}^{f_{\omega+1}(3).2}(3) >> f_{\omega+1}^2(3)\) then \(D(2,0,0) >> f_{\omega+1}^2(3)\) \(D(2,0,1) >> f_{\omega}^{m+(m+2).(m+1)/2}(3)\) where \(m >> f_{\omega+1}^2(3)\) then \(D(2,0,1) >> f_{\omega+1}^3(3) = f_{\omega+2}(3)\) \(D(2,0,2) >> f_{\omega}^{m.2}(3)\) where \(m >> f_{\omega+2}(3)\) then \(D(2,0,2) >> f_{\omega+2}^2(3)\) \(D(2,0,3) >> f_{\omega+2}^3(3) = f_{\omega+3}(3) = f_{\omega.2}(3)\) \(D(2,0,n) >> f_{\omega.int((n+9)/6))+mod_3(int((n+4)/2))}^{mod_2(n+1)+1}(3)\) where \(int(n)\) is the integer part of n and \(mod_x(n)\) is the value of n modulo x e.g. \(D(2,0,9) >> f_{\omega.3+0}^{1}(3) = f_{\omega^2}(3)\) e.g. \(D(2,0,100) >> f_{\omega.18+1}^{2}(3) = f_{\omega^2+\omega.15+1}^{2}(3)\) need to check this D function examples with 3 parameters - continues for \(D(2,m,n)\) \(D(2,1,0) = D(1,D(2,0,1),D(2,0,1)) = D(2,0,2) >> f_{\omega+2}^2(3)\) \(D(2,1,1) = D(1,D(2,1,0),D(2,1,0)) = D(2,0,3) >> f_{\omega+3}(3) >> f_{\omega.2}(3)\) \(D(2,1,n) = D(2,0,n+2)\) \(D(2,2,0) = D(1,D(2,1,2),D(2,1,2)) = D(2,1,3) = D(2,0,5) >> f_{\omega.2+1}(3)\) \(D(2,2,n) = D(2,1,n+3) = D(2,0,n+5)\) \(D(2,3,n) = D(2,2,n+4) = D(2,1,n+7) = D(2,0,n+9)\) \(D(2,m,n) = D(2,m-1,n+m+1) = ... = D(2,0,n-1+(m+2).(m+1)/2)\) e.g. \(D(2,3,3) = D(2,0,3-1+(3+2).(3+1)/2) = D(2,0,12) >> f_{\omega.int((12+9)/6))+mod_3(int((12+4)/2))}^{mod_2(12+1)+1}(3)\) \(= f_{\omega.int(21/6))+mod_3(int(16/2))}^{mod_2(13)+1}(3) = f_{\omega.3+mod_3(8)}^{1+1}(3) = f_{\omega.3+2}^{2}(3)\) e.g. \(D(2,12,10) = D(2,6,73) = D(2,3,91) = D(2,0,100) >> f_{\omega.18+1}^{2}(3)\) then \(D(3,0,0) = D(2,D(2,3,3),D(2,3,3))\) \(= D(2,0,m-1+(m+2).(m+1)/2)\) where \(m = D(2,3,3) >> f_{\omega.3+2}^{2}(3)\) \(= D(2,0,m-1+m^2/2+m.3/2+1) = D(2,0,(m^2+m.5)/2) >> f_{\omega.f_{\omega.3+2}(3)}(3)\) need to check this \(D(3,0,1) >> f_{\omega.f_{\omega.f_{\omega.3+2}(3)}(3)}(3)\) need to check this \(D(3,0,2) >>\) need to check this \(D(3,0,n) >>\) need to check this \(D(3,1,0) = D(2,D(3,0,1),D(3,0,1)) = D(3,0,2)\) \(D(3,1,1) = D(2,D(3,1,0),D(3,1,0)) = D(3,0,3)\) \(D(3,1,n) = D(3,0,n+2)\) \(D(3,2,n) = D(3,0,n+5)\) \(D(3,m,n) = D(3,0,n-1+(m+2).(m+1)/2)\) Recap and Speculating ahead a little \(D(3, 9) >> 1,000,000\) \(D(3, 206) >>\) Googol \(D(4,2) >>\) Googolplex \(D(1,0,0) >> f_{\omega}(3)\) \(D(6,1) >> g_1\) where \(g_64 = G\) Graham's number \(D(1,1,0) = D(1,0,2) >> f_{\omega+1}(3)\) \(D(1,9,9) = D(1,6,36) = D(1,3,54) = D(1,0,63) >> g_64 = G\) Graham's number \(D(2,0,1) >> f_{\omega+2}(3)\) \(D(2,0,3) >> f_{\omega.2}(3)\) \(D(2,0,9) >> f_{\omega^2}(3)\) \(D(4,0,2) >> f_{\omega^{\omega}}(3)\) need to check this \(D(l,m,n)\) has a growth rate of \(f_{\epsilon_0}(n)\) need to check this Next My next blog post will introduce a new Alpha function that I have been thinking about. Alternative examples for \(D(2,0,n)\) functions \(D(2,0,0)\) grows significantly faster \(= D(1,D(1,2,2),D(1,2,2))\) Using this formula 12 \(f_b^{f_{b+1}(a)+a}(a) = f_{b+1}^2(a)\) we get \(f_{\omega}^{f_{\omega+1}(3)+3}(3) >> f_{\omega+1}^2(3)\) then \(D(2,0,0) >> f_{\omega+1}^2(3)\) \(D(2,0,1) >> f_{\omega}^{m+(m+2).(m+1)/2}(3)\) where \(m >> f_{\omega+1}^2(3)\) Using this formula 15 \(f_{b}^{f_{b+1}^{a-1}(a).2}(a) >> f_{b+2}(a)\) we get \(f_{\omega}^{f_{\omega+1}^2(3).2}(3) >> f_{\omega+2}(3)\) then \(D(2,0,1) >> f_{\omega+2}(3)\) \(D(2,0,2) >> f_{\omega}^{m.2}(3)\) where \(m >> f_{\omega+2}(3)\) Using this formula 14 \(f_{b}^{f_{b+1}^n(a).2}(a) >> f_{b+1}^{n+1}(a)\) we get \(f_{\omega}^{f_{\omega+2}(3).2}(3) >> f_{\omega+2}^2(3)\) then \(D(2,0,2) >> f_{\omega+2}^2(3)\)